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(a)
To Find: The period and frequency.
(a)
Answer to Problem 23P
The period of the motion is 0.49 s and the frequency of the motion is 2.04 Hz .
Explanation of Solution
Given:
Mass, m = 0.755 kg
Spring constant, k = 124 N/m
Initial speed, vi = 2.96 m/s = vmax
Formula used:
At t = 0, time period (T) is obtained by the following formula,
T = 2π×√mk.................(1)
Here m is the mass of the spring,
k is the stiffness of the spring,
Write the expression for the frequency.
f = 1T.................(2)
Here, T is the time period of the oscillation of spring.
Calculation:
Substitute the given values in the above equation,
T = 2π×√0.755124=0.49 s f=10.49=2.04 Hz
Conclusion:
Thus, the period of the motion is 0.49 s and the frequency of the motion is 2.04 Hz .
(b)
To Find: The amplitude.
(b)
Answer to Problem 23P
The amplitude of the motion is 0.231 m
Explanation of Solution
Given:
Mass, m = 0.755 kg
Spring constant, k = 124 N/m
Initial speed, vi = 2.96 m/s = Maximum velocity vmax
Formula used:
Amplitude of the motion is determined by the following formula,
Vmax = ω × A.................(3)
Here, Vmax is the maximum velocity,
ω = √km.................(4)
Here m is the mass of the spring,
k is the stiffness of the spring,
From equation 3 and equation 4.
Vmax = √km × A
Calculation:
Substitute the given values in the above equation,
Vmax = √km × AA = Vmax × √mk = 2.96 × √0.755124 = 0.231 m
Conclusion:
Thus, the amplitude of the motion is 0.231 m .
(c)
To Find: The maximum acceleration.
(c)
Answer to Problem 23P
The maximum acceleration is 37.9 m/s2 .
Explanation of Solution
Given:
Mass m = 0.755 kg
Amplitude A = 0.231 m
Spring constant k = 124 N/m
Formula used:
The maximum acceleration is determined by the following formula,
amax = A × ω2.................(5)
Here A is the amplitude of the vibration in the spring,
ω is the angular velocity.
ω = √km.................(6)
Here, m is the mass of the spring,
k is the stiffness of the spring,
On comparing equation (5) and equation (6)
amax = A ×kmCalculation:
Substitute the given values in the above equation,
amax = A × km = 0.231 × 1240.755 = 37.9 m/s2
Conclusion:
Thus, the maximum acceleration is 37.9 m/s2
(d)
To Find: The position as a function of time.
(d)
Answer to Problem 23P
The position as a function of time is x = 0.231 sin (12.82)t .
Explanation of Solution
Given:
Amplitude A = 0.231 m
Frequency f = 2.04 Hz
Formula used:
Mass started at equilibrium position at x = 0, so the displacement (position) function with respect to time is given by sine function.
x = A sin (ωt)
Here, A is the amplitude of the vibration,
ω is the
t is the time period.
ω = 2πf
f is the frequency of the vibration.
Calculation:
Substitute the given values,
x = A sin (ωt)ω = 2πfx = 0.231 sin (2π×2.04×t)x = 0.231 sin (12.82)t
Conclusion:
Thus, equation of displacement as a function of time is x = 0.231 sin (12.82)t
(e)
To Find: The total energy.
(e)
Answer to Problem 23P
The total energy of the object is 3.31 J
Explanation of Solution
Given:
Maximum velocity Vmax = 2.96 m/s
Mass m = 0.755 kg
Formula used:
The kinetic energy of an object is the total energy of an object at an equilibrium position, which is given by following formula,
Emax = 12×m×vmax2
Here, vmax is the maximum velocity,
m is the mass of the object.
Calculation:
Substitute the given values,
Emax = 12 × 0.755 × (2.96)2 = 3.31 J
Conclusion:
Thus, the total energy of the object is 3.31 J .
Chapter 11 Solutions
Physics: Principles with Applications
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